Similar questions have been asked here, but I'm not able to find this exact question. For a European call option, if it's very in-the-money, wouldn't a higher volatility decrease the probability it finishes in-the-money? But if it had like a 0 volatility, wouldn't the chance it would finish out-the-money be 0 percent?

3 Answers 3


Volatility is a measure of the expected price variance over time, up or down. Increased volatility means a wider range of expected prices, so the ITM call option is seen as both more likely to expire OTM, and more likely to expire even deeper ITM. The extrinsic value in options is basically the price of uncertainty, so greater uncertainty means greater extrinsic value.


The answer by @Hart CO is intuitively correct but within a Black Scholes framework, this is actually quite a brain teaser as the probability of ITM options expiring ITM indeed decreases with increasing vol.

Intuitively, volatility of an underlying doesn't care about options. If you think about it, why would an OTM option benefit from vol if an ITM would not. The former still needs prices to increase, whereas the latter is already ITM. There is only one underlying (for any given option chain on some security). For calls, if we believe the higher strikes benefit, surely the lower strikes must benefit as well.

Mathematically, this is a lot more nuanced because the higher vol, the more the lognormal distribution tries to extend itself on both sides of the definition domain where it hits a boundary at zero, and probability accumulates. For this reason, the risk neutral probability of ending up in the money in a Black Scholes setting does indeed decrease all the time for ITM calls when IV increases.

For OTM calls on the other hand, increasing IV initially increases the probability, before it also starts to decrease. The explanation pins down to the so called volatility tax and convexity adjustment. Probably even more interesting, although the risk neutral probability of the call option expiring ITM becomes zero for very high vol, the price itself approaches its maximum possible value (the discounted forward / spot in case of no dividends).

Since mathjax isn't supported here, I recommend looking at this answer on quant stack exchange. It combines a mathematical explanation with animated GIFs and charts to explain this in detail. Just a quick teaser below:

enter image description here

Interesting side remark; the same logic applies to longer time periods. Ultimately, limits can be a strange thing. Warren Buffett explained his take on the Black Scholes formula for long-dated options in his 2008 letter to the Shareholders of Berkshire Hathaway.

"The Black-Scholes formula has approached the status of holy writ in finance, and we use it when valuing our equity put options for financial statement purposes. ... If the formula is applied to extended time periods, however, it can produce absurd results. In fairness, Black and Scholes almost certainly understood this point well."

Last but not least, it's questions like this that sparked my interest in quantitative finance and why I ended up working as a derivatives quant.


Let’s say the strike price is $20, current price is $30, and little volatility. The expected change is $0, with a small variance, so the option is worth $10: You will get $10, or two dollars more, or two dollars less.

But if the volatility is high: If the shares go up from $30 to $50, you get $30 instead of $10 when you exercise the option. But if the price goes from $30 to $10, you don’t exercise the option and lose only $10, not $20.

Your losses stop when the share price meets the strike price. So with high volatility, your gain if the stock goes up is more than your loss if it goes down. So the option is worth more.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .